how to automatically close tinyweb console after running from batch file

Question

I'm using TinyWeb server. I run it using a batch file START_tiny.bat:

c:
cd\
cd tiny
cd bin
tiny c:\tiny\root 8080
exit

The problem is that after tiny.exe is executed, the console won't close so it hangs here: I can manually close the console window, and it will continue to run as expected, but I was just wondering if there was a way in the batch file to make sure it closes after the program is invoked.

EDIT: the solution was:

cmd /c start tiny c:\tiny\root 8080

Try using cmd /C start program_here. You might be able to isolate the initiation of it — IoCalisto, Aug 08 '19 at 18:31
You can change the first four lines to this: `CD /D "C:\tiny\bin"`. — Compo, Aug 08 '19 at 18:33

Mofi · Accepted Answer · 2023-08-12T14:34:31.617

The Windows Command Processor cmd.exe halts processing of the batch file as long as tiny.exe is running and waits for its termination, even if this application is a GUI application not opening any window or opening a GUI window.

The command exit is not necessary at all and is just bad for debugging the batch file.

It would be better to use a shortcut file (*.lnk) to just start TinyWeb server. Target in properties of shortcut file would be C:\tiny\bin\tiny.exe C:\tiny\root 8080 and Start in would be C:\tiny\bin or C:\tiny\root or whatever should be the current directory on starting TinyWeb. No console window is shown on tiny.exe not being a console application which I don't know as not having downloaded and installed this application.

The command start can be used in a batch file for starting TinyWeb server as separate process by cmd.exe without waiting for its termination.

One of the following command lines could be used in the batch file:

start C:\tiny\bin\tiny.exe C:\tiny\root 8080
start /DC:\tiny\bin tiny.exe C:\tiny\root 8080
start /D C:\tiny\bin tiny.exe C:\tiny\root 8080
start "TinyWeb server" C:\tiny\bin\tiny.exe C:\tiny\root 8080
start "TinyWeb server" "C:\tiny\bin\tiny.exe" "C:\tiny\root" 8080
start "TinyWeb server" /D"C:\tiny\bin" tiny.exe "C:\tiny\root" 8080
start "TinyWeb server" /D "C:\tiny\bin" tiny.exe "C:\tiny\root" 8080

Open a command prompt, run start /? and read the output help for an explanation of the above command lines.

The parameter /D specifies the start in respectively current directory for the started executable. It can be specified immediately after /D or separated with a space from the switch.

Command start interprets first string in quotes as window title. For that reason it is necessary to explicitly specify a window title in double quotes on one of the other arguments is enclosed in double quotes even if no console window is opened because of started application is a Windows GUI application. An empty title string specified with just "" is enough for GUI applications started by command start as separate process.

The command line cd /D "C:\tiny\bin" can be used to change current directory to C:\tiny\bin even on current directory being on a different drive. The help output on running in a cmd window cd /? explains the parameter /D to change also drive if necessary and not only current directory on current drive.

how to automatically close tinyweb console after running from batch file

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